Chapter 2 System Ist Order Response

8/22/2019 Chapter 2 System Ist Order Response

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Chapter 2 (page#119)

Continuous-TimeSystem Responses

2.2 Response of Ist Order System

2.3 Response of Second-Order Systems

2.5 Stability Testing

Introduction

2.4 Higher- Order System Response


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Introduction

Chapter: Continuous-Time System Responses

Higher Order Systems = Sum of first order and second order systems. Why?

Time function f(t) = u(t); Laplace Transform F(s) = 1/s

[see table B.1 P=836]

Examples of Step Inputs:

Common input to control system is the : Step function


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OtherTypes of Inputs:

Ramp

Sinusoidal

Pulse

Repeated Sequence

Random Number

In automatic Landing of

aircraft, the aircraft is

commanded to follow for the

glide slop; this glide slop is

commonly approximately 3o

For investigation: response of Ist & IInd order systems are

checked for various inputs


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0

0

as

b

)s(R

)s(C)s(G)s(T

2.2(p 120) RESPONSE OF FIRST-ORDER SYSTEMS

1

s

K

)s(R

)s(C

Kb

aLet

0

0

&

1

K)s(R)s)(s(C 1

K)s(R)s()s(C

1

K)s(R)s()s(C

11

Can be written as

= time constant:

= RC (seconds); = L / R (sec)

K=DC gain i.e steady state valueG(0)=K

)......(rbca

dt

dc1200 )s(Rb]aS)[s(C 00

/s

/K

)s(R

)s(C

1

X

S = -a0

Pole =1/=-a0

X

S = -a0

Pole =1/=-a0

CharacteristicsEquation: Why?

C(t)

t

K

-S0

S=0-Transient

-Natural Steady-state

Ts=4


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K)s(R)s()s(C

11

The inverse Laplace Transform (Time domain)

K)s(r)dt

d()t(c

11K)s(r

)t(c

dt

)t(dc

1

Now take the Laplace transform and include the initial conditions

)s(Rk)s(C

)(c)s(CS

0

ILLT of a constant = impulse response; impulse function c(0)(t)

C(0) : initial condition and is a constant value

2.2(p 120) RESPONSE OF FIRST-ORDER SYSTEMS .Cont


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)s(R

s

/K

s

)(C)s(C

11

0

)s(RK)s(C

)(c)s(CS 0

)s(R

K

)(C)s)(s(C 01



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)s(R

s

/K

s

)(C)s(C

11

0

C(s)

1

1

s

1s

/K

+

+

R(s)

c(0)

1s

/K

R(s) C(s)C(0) initial condition

is impulse input


If initialconditions=0


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Unit-step response for first-order system =R(s) = A/s

)s(Rs

K

)s(C

1

ss

/K)s(C1

1

)s(R)s(G)s(C

1

s

K

s

K

1

s

/KR(s) C(s)

= 1/s



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1s

K

s

K)s(C

Inverse Laplace transform [see table B.1 P=836]

01 t,)e(K)t(ct

t

KeK)t(c

Originates in the pole of R(s), has

Constant Value & is Called:

Forced Response

Steady State Response

Originates in the System G(s) Transfer

Function & is Called:

Transient Response

Natural Response

As t0 Ke-t/

goes to zero

e-50

TABLE 4.1

t e-t/

0 1

0.3679

2 0.1353

3 0.0498

4 0.0183

5 0.0067

TABLE 4.1

t e-t/

0 1

0.3679

2 0.1353

3 0.0498

4 0.0183

5 0.0067



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t e-t/

0 1

0.3679

2 0.1353

3 0.0498

4 0.0183

5

0.0067

)e(Kt

1

tKe

Figure 2.1&2.2 Step response of first order system

Slop = K/ If decay at its initial rate, it wouldreach a value of zero in sec

2 %

1 %

Settling

Time

Ts=4

-K

K

-0.5K

0.5K

0t

C(t)

Ts=4



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Example Find the unit step response of a system with the transfer function

150

52

s.

.

)s(R

)s(C)s(G

2

5

sX

S = -2

Pole =1/

ss)s(R)s(G)s(C

1

2

5

2

2525

s

/

s

/

)e()/()t(c t2125 )e(Kt

1

K= 5/2 =steady state response or forced response

s=-2 s= 1/ =2 =0.5 sec Ts=4 =2 sec

Characteristic equation ? S+2=0

ponsenaturalres)e)/( t225 t Will go to zero (decay exponentially)Ts=4 =2 sec

C(t)

t

2.5

5.0;5.2;1

)(

Ks

KsG

1

/

s

K

Step response R(s) = A/s = 1/s unit step response


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150

52

s.

.

)s(R

)s(C)s(G 5052

1.;.K;

s

K

)e(.)t(c

Similarly

t

152

S=d/dt=0

t > 4S=d/dt 0t < 4

52

1050

520

0

150

52

.

x.

.)(G

s

s.

.

)s(G

5215244

.)e(.)t(ct

TABLE 4.1

t e-t/

0 1

0.3679

2 0.1353

3 0.0498

4 0.0183

5 0.0067

TABLE 4.1

t e-t/

0 1

0.3679

2 0.1353

3 0.0498

4 0.0183

5 0.0067

521050

520

0

.x.

.)(G

s

C(t)

t

2.5

Ts=4 =2 sec

K

015200

)e(.)(ct

015200

)e(.)(ct


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Example was stable: (Pole in left half of S-plane) the forced response is the steady-state response and the natural response is the transient response

DC gain:

If the system is stable, so that c(t) has a (dc) steady-statevalue:

For unstable responses, "steady-state" and "transient"

are meaningless

XS = -2

Pole =1/

2

5

s

)s(T


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MATLAB m.File Program

num=[0 0 5];

den=[1 2 0];

[r,p,k]=residue(num,den)

pause

G=tf ([0 5],[1,2]);

Step (G)

>>r =

-2.5000

2.5000

>>p =

-2

0

>>k =[ ]

t = 0:0.01:3;

c=2.5*(1-exp(-2*t));

plot(t,c)

MATLAB Si li k


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MATLAB Simulink

D2 1 F D ill P bl S l ti (P 123)


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0

0

)(

)()()(

as

b

sR

sYsGsT

D2.1 For Drill Problems Solutions (Page 123)

ConditionsInitialNotuea

Ab

a

Abty ta )()( 0

0

0

0

0

00

0 )0()()(as

YsR

as

bsY

Zero-state

componentZero-input

component

ICWithtuea

Aby

a

Abty

ta)()0()( 0

0

0

0

0

10)0(

);(6)(

;3

3

)(

)(Pr

y

tutr

ssT

aoblemDrill


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dt

diLiRV

)....(rbcadt

dc1200 ViR

dt

diL

)s(V)RLS)(s(I

)RLS()s(V

)s(I

1

1

s

K

)s(R

)s(C

)L/RS(

L/

)s(V

)s(I

1

/s

/K

)s(R

)s(C

1


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X

S = -a0

Pole =1/=-a0

Characteristics

Equation

X

S = -a0

Pole =1/=-a0X

S = -a0

Pole =1/=-a0

Chapter 2 System Ist Order Response

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